AIEEE
FEN
Test Assessment, Analysis & Solutions Some students are of the opinion that for AIEEE, practice is everything. At PAGE we recognize that practice is one of the most important constituents of a good study plan. However, it is not enough to blindly attempt tests. After every test, one needs to spend time reviewing it. The following charts will give you a place to start your test review First check the question paper with answer keys to find out marks scored. Religiously note down the number of questions that you obtained correct, wrong or left unanswered in various section types. Also make a subjective self-judgment ascertaining the cause of your performance in the various sections. Summary performance table.
PAGE Physics
Total number of questions Questions Attempted Correct Answers
Chemistry
Maths
Total
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Wrong Answers Total Scored
Analysis of wrong questions: Reasons for wrong questions (To be filled after you have attempted wrong questions on your own after the test)
No. Questions
A. Know the solution / answer for the question but failed to get it right in the exam: Calculation mistake
B. Know the solution / answer for the question but failed to get it right in the exam: Applied wrong concept
C.
Do not know the concept and couldn’t solve even after exam
Total Number of questions attempted wrong If score of A &B is high then you need more practice and read questions more carefully. If C is very high you need to revisit the topic in the textbook.
PAGE PAGE
If the number of easy, average question not attempted by you is high then you need to focus on selection of questions. In AIEEE you should select questions you can solve fast and leave lengthy and difficult questions.
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1
SOLUTIONS
1.c.
CODE : F E N
Heat Absorbed Entropy = Temperature 8.a.
Q ⇒S= T
energy charged particles, like α - particle, proton etc.
⇒ [S] = ML2 T −2 K −1 9.b.
Also,
1 E = kaT 2 2 −2 E ML T ⇒ [k B ] = = = ML2 T 2 K −1 K T
2.a. 3.d
analogous to electrical coductivity σ since σ decreases with temperature rise for the same reason and so does K. Cosmic rays are coming from outer space, having high
β - rays are stream of high energy electrons, coming from the nucleus of radioactive atoms. The potential difference accrose the capacitor remais E' Current through the battery naver becomes zero be cause the resistor provides and alternative path. The rsistor dissipates energy as heat. Under revrse basic majority carriers are drawn away from junction. This actoinwidens depletion region and increases barrier poteantial.
PAGE 10.a.
on rotating themagnet, no chang influx is linkedwiththe coil. Therefore, induced e.m.f/currentis zero As the block slowly compress the spring its kinethic energy remains constant i.e. zero gravitational potential energy decvreasesand spring potential energy increases. The rate of decrease in gravitationa potential energy is ore then the rate of increase of spring poitential energy because the external force does work on the block The action of a nicol prism is based on double refraction and dichroism. When a ray of light enters into a calcite crystal, it is split into two rays O and E. Both are plane polarised in mutually perpendicular planes. One of them suffers total reflection and absorption while the other comes out as plane polarised. Selective absorption is called dichroism. If the trolley does not move backward, then the momentum will not be conserved with respect to ground. The distance moved by the trolley depends only on the distance travelled by the man irrespectie of the velocity of man.
11.c.
From s = ut +
1 2 1 2 1 at = at = (µ g) t 2 2 2 2
2s 1 ∴t ∝ µg µ
t=
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4.d.
5.d.
Since the distance travelled by trolley is
12.a.
13.a.
7.d.
= 2m v sin 45° = 2 mv × 1/ 2 = 2 m.v. It is a balanced stone bridge and so the R is not in the circuit. In given figure resistances R1 and R2 are in series. Also resistances R3 and R4 are in series. Now equivalent of R1 and R2 is in
mL m+M
R 1 = 10Ω
m <1 therefore ,it canot exceed L because m+M
6.d.
Resolving momentum of projectileintotwo rectangular components at starting point and arriving point We note that the horizontal components of momentum remains unchanged but the vertical components of momentum.mv sin 45° each act in opposite directoin. Therefore change in momentum = mv. sin45° -(-m v sin 45°)
A
When a system is in equlibrium under the action of three forces whose lines of action are not parallel, their lines ofaction must be concurrent. When the temperature of copper creases, the amplitude of vibrations of the lattice atos at the lattice sites in copper is also increased. Thus, conduction electrons, which is the means of transport heat in copper, collide more often with the copper atoms. The mean time between collision is decreased and on the average, the man speed of the electronsat which the heat is trans ported is thus reduced. As a result, the thermal coductivity of copper devreases with temperature rise of copper. Note: For copper (metal) thermal conductivity K is
R 3 = 10Ω
PAGE
R 2 = 10Ω
B
R R 4 = 10Ω
parallal to equivalent of R3 and R4 . Therefore the given circuit can be simplified as
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energy only. Initil internal energy of the gas is R1 =10Ω
5 U 1 = N R T 2
R2 =10Ω
A
B
Sincen moles get dissociated into atoms, therefore, after heating, vessel contains (N-n) moles of di atomic gas and 2n moles of a mono-atomic gas. Hence theinternal energy of the gas, after heating, will be equal to
R4 =10Ω
R3 =10Ω
20Ω
5 5 U 2 = ( N - n) R T + 2m R T 2 2 =
20Ω
Hence, the heat supplied = increase in internal energy
10Ω
14.b.
m (l + x) ω = kx;∴ x =
15.b. 16.c. 17.c.
PAGE
1 = (U2 − U1 ) = nRT 2
Hence we get equivalent resistance of circuit as 10Ω Let x be increase in length of the spring. The particles would move ina circular path of radius. (l+x). Centripital force = force due to the spirng. 2
x = (t = 3) 2 , =
m ω2 l
23.b.
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k − m ω2
24.a.
dx = 2(t + 3) or v ∝ t dt
Resistance of 40 W bulb = =
Resistance of 60 W bulb =
Restoring force = − Mg sin θ, Resistance of 100 W bulb =
Mg g x or a = − x 4r 4r
1 q Electric intensity = 4 π ε × 2 0 r or q max =
1 9 × 109
r 2E =
1 9 × 109
or q max = 2 × 10−3 coulomb
200 × 200 = 666.7 Ω 60 200 × 200 100
= 400 Ω
Hence choice (3) is correct. Since the bulbs are connected in series, the current in each is the same. Hence Choice (2) is incorrect. The toral resistance = 1000 + 666.7+400 = 2066.7 Ω . Therefore, the current in the circuit is
4r r = 4π g g
And T = 2 π
v 2 200×200 = R 40
1000 Ω
When observed through the plane surface, it behaves as
That is Ma = −
20.c. 21.b. 22.b.
m v sin θ 2π m ;p = cosθ qB B
or θ = tan -1 2π
x x = − Mg = − Mg OG 5r − r
19.b.
r=
since r = p , therefore, tan θ = 2π
a slab. Hence µ = 3/2 = 1.5 18.b.
5 1 NRT + nRT 2 2
I=
200 = 0.097 A 2066.7
Hence the choice (4) is also incorrect. Since the 40 W bulb has the highest resistance and the current in each bulb is the same, the potential difference across the 40 W bulb is the highest. Hence the correct choices are (1) and (3) In the absence of a magnetic field, the particle will experience gravitational force mg. As a result the particle will not continue moving in the horizontal direction but will describe a parabolic path. So a magnetic field must be present and its direction must be perpendicular to the direction of the velocity.; The magnetic force experi-
PAGE
× 2.5 × 2.5 × 3 × 106
25.b.
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Since the gas is enclosed in a vessel, therefore, during heating process, volume of the gas remains constant. Hence, no work is done by the gas is used to increase its internal
enced by the particle is given by
F = q(v×B)
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F = q υ B sin θ . If
33.b.
the particles is to move in the horizontal direction, this force must balance the force of gravity, i.e.,
34.a.
The magnitude of the force is
mg = q υ B sin θ
0 = V − y(t′ − t ′) = xt ′ − y(t − t ′)
The minimum value of B corresponds to
∴ t′ =
sin θ = 1 or θ = 90° . Thus mg = q υ B or mg 0.5×10−3×9.8 = = 3.27 T qυ 2.5×10−8 ×6×104
B=
As optical path SB of lower slit is increased therefore, fringe pattern shifts somwwhat downwards. Velocity of car after time t’ V = 0 + xt’ If it decelerates for (t-t’) finally comes to rest then
35.b.
y xy t ∴ V = xt′ = t x+y x+y
For student 320 =
1 × 10 × t 2 t = 2
Hence the correct choices are (1) and (3). 26.b.
For superman 320 = u (t − 5) +
E I = 0 (1 − e −t / τ) R
27.c.
5L 3R
PAGE
according to ideal gas equation, PV = RT
⇒
From the given relation,
K.E. of the block at P = 1.5 mg. This is wasted away in doing work on the rough flat part,
P
37.
Q
1.5 m
∴ The block comes to rest at mid-point of PQ K= 2
1 2 Iω and 2
I = (M /12) + M( /2) ∴ I=
2
M 2 ; 3
38.
⇒
T2 27 = 300 8
⇒
T2 = 300 ×
⇒
T2 = 675 K or 402°C
2/3
9 4
b)The termial p.d of battery V = E - ir V = 12 - 60 x 5 x 10 - 2 V = 12 - 3 ⇒
N
V = 9 volt
Ip
s b)We have, N = I p s
⇒
25 I p = 1 2
⇒
Current in primary = 50 A
V b) P 8
4/3
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5 / 3−1
⇒
PAGE
1 M 2 2 1 2 2 = K ω = M ω ∴ 2 3 6
γ −1
T2 V1 = 300 8 / 27V1
⇒ ⇒
1.5 = 7.5 m. ∴ 1.5mg = µ mg.x or x = µ
K.E.
T1 V1γ- 1 = T2 V2γ - 1
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mv v v = = 2qd 2(q/m)d 2Sd
15 m
31.c.
RT1 γ RT2 γ .V1 = .V2 V1 V2
T2 V1 = T1 V2
1 (L) n(x 2 − x1 ) L2 T = = [L2 T −1 ] D= n 2 − n1 1/L3 29.d. 30.b.
In adiabatic expansion, P1V1 γ = P2 V2 γ
⇒
1 2 i.e. mv = Bqv × d or 2
28.d.
36.b.
(5L) 3
The charged particle will not hit the wall if the K.E. of charnged particle is totally spent against the retarding force due to magnetic field while travelling a distance.
B=
1 × 10 (t − 5) 2 2
Find u = 91.6 m/s
R
τ=
320 5
⇒
Ip = 50A
4/3 4/3 = P [V] or P [8] or 0 0
P = 16P 0
32.b.
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40.d. 41.b.
Since pulley B is in static equilibrium threfore, tension in thread passing over pulley A is equal to 2T. But it
The echo is not heard distinctly when the echo and the next beat fall on the ear simultaneously. That is time per beat = time taken by the reflected beat to reach the man. Hence
2 d 60 3 = = c 40 2
And
balances block m1
2(d − 80) c
m 2 + m3 4 = m 2m3 m1
or
This gives d = 240 m 42.b.
4 m 2 m3 g m 2 + m3
Hence, m1g = 2T =
Time taken for swimming across in shortest time is
t1 = L / v and that for swimming across over the shortest distance is t 2 = L / v sin θ. Hence
50.c.
λ = h/mv = h / 2meV
51.b.
Given, half life of radioactive element, T1/2 = 12.5 hour then decay constant, λ =
t1 / t = sin θ.
PAGE ⇒
L
⇒
θ 43.b.
According to the Lenz’s law the induced current opposes the cause that produces it. 44.c. Since the two cars complete circle in the same time, their angular velocities must be the same. 45.c. The reading of balance A will decreased due to the upward thrust caused by buoyance. The upthrust will be equal to the weight of water displaced. The net downward force due to mass immersed in water wilol add to effective weight of the system. So thereadig of ablance B will increase. 46.d. Let y = A sin ωt
λ=
0.6931 T1 / 2
0.6931 12.5
λ = 5.54 x 10 - 2 per hour
Now we have the formula
N = N 0 e − λt
where N0 → the quantity of radioactive element at t=0 and N → the remaining quantity of radioactive element at time t.
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For y = A/2, we find sin ωt = or ωt = sin
47. 48.
−1
⇒
1 = 256e − λt
⇒
e − λt = 1 / 256
taking log of both sides, we have −λt = −8 log e 2
1 2
1 2
or ωt = π/6 or 5π/6 So the phase difference = 5π/6 - π/6 = 2 π/3 c)On closing one of the slits in young’s double slits experiments uniform illumination is obtained.
⇒
λt = 8 × 0.6931
⇒
t=
8 × 0.6931 5.54 × 10 −2
⇒
t=
5.54 × 10 2 5.54
⇒
t = 100 hours
52.b.
d)Fuse wire should be such that it melts immedi ately when strong current flows through the curcuit. The same is possible if its melting point is low and resistivity is high.
49.b.
PAGE
If m1 isat rest then pulley B isalso at rest. In that case,
In this figure the current flows in loop (A) because the other diode is reverse-biased in loop (B) Then by applying Kirchoff’s law in loop A 20I + 30I = 5
magnitude of accelerations of m 2 and m 3 will be equal
m2 − m3 g and tension in the thread connected m2 + m3
to
⇒ ⇒
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between m 2 and m3 will be equal to
T=
2 m 2 m3 g m 2 + m3 .
53.b.
50I = 5 I = 5/50 amp.
At one ofthe extreme position, wt of block = re storing force. At the other extreme position wt of block and restoring force both act downward directions. So
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54.b.
thewt of block there is double thants weight. Applying Boyle’s law, we have
Mol, formula = n × EF . and n =
62.c.
Mol of O 2 =
63.c.
Hence, mω 2 y = mg or
64.b. 65.b. 66.c
Not more than two electrons can be present insame atoic orbital, this is Paulis exclusion principle. Non electrolytes do not undergo dissociation Energy rich states lead to instability A buffer olution does not chage itspH
y = g/ω2 9 + 8(2 π )2 = 0.25m
67.b.
C 2 H 5 OH is a non-electrolyte
Given that | A + B |= A − B Squarig both sides, we get
68.c.
Boron compounds are electron deficient and therefore. behave as Lewis acids.
2 x x = + T T T
PV PV PV PV or + =x +x T T T T
xT 2−x Weight kept on thesystem will seperate from thepiston when the maximum force just exceeds the weight of the body.
or T = 55.b.
56.d.
| A + B |2 = ( A − B) 2 or
(
of molecules of SO 2 will be less then O 2 .
)
AA + AB + BA + BB = A 2 + B2 − 2AB or A 2 + 2A.B + B2 A 2 + B2 − 2AB
69.a.
The conjugate acid of NH−2 isweakly basic
70.
a) Non-metals have electronegative nature, while metals have electropositive nature. So, the group which has maximum capacity to form anio is N, O, F.
71.a. 72.c.
∵ AA = A and A.B = B.A 2
(∵ AB = AB cos θ)
73.a.
cos θ - 1 ⇒ θ = 180° Change in stopping potential = Vs2 − Vs1
=
58.c.
74.a.
10 × 3 × 10 10 − −19 3000 4000 1.6 × 10
6.6 × 10
8
10
10
75.a.
HCl
AuCl 3 + HCl → H [AuCl 4 ] which is used for toning in photographs
4 qV π r 3pg = or V ∝ r 3 3 d
60.a.
3 r 2 ∴ V2 = V1 2 = 400 × = 3200V 1 r1 In going from A to C ia B,
∆Q = 600 + 200 = 800J
PAGE 76.d.
70 kJ
ENERGY
3
∆W = WAB + WBC = P × O + Pa BC) = 8 × 104 × (VD - VA )
Au is not affected by acids and alkalies. The compound
AuCl 3 X is AuCl 3 which forms a complex with conc.
B2 n 2 r1 2 r = × = × = 4 or B = 4B 2 1 B1 n1 r2 1 r/2 mg = q E or
Gypsum (CaSO 4 2H 2 O) contains a lower % of calcium than plaster of pairs (CaSO 41/2H 2 O)
µ o 2 π ni n As B = 4 π r ; so B ∝ r thus
59.c.
The calculatedvalue of O.N. of Na 2S4O6 is2+4x+6(-2) = 0 or 2+4x-12 = 0 or x = 2.5. This is the average of two values. 2 × ( +5) + 2 × 0 10 + 0 = = 2.5 4 4
he 1 1 − e λ 2 λ1 −34
(i) Nuclei with pandn even (charge and mass even) have zero spink (eg., helium - 4). (ii) Nuclei with p andn odd (charge odd andmass even) have integral spin (e.g., deuterium). (iii) Nuclei withodd mass have half-integral spin (e.g., hy drogen)
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or 2AB cos θ = - 2AB
=
6 6 & mol. of SO 2 = the number 32 64
PAGE
(A + B) 2 = (A − B)2 ∵| A |= (A) 2
57.b.
Mol.wt EF.Wt
61.b.
A+B
100 kJ
30 kJ C
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PROGORESS OF REACTION
= 8 × 104 × (5 × 10-3 - 2 × 10-3 ) = 240 J ∆u = ∆Q − ∆w = 800 − 240 = 560 J
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77.a.
A portion of the spiral chain in K[Cu(CN)2 ] is Cu
87.a.
Glycine (H2 N.CH2CO2H) containsanacidicgroup as well as a basic group.
88.a. Cu
C N
Cu
C
N
89.b.
Solution gives 1.1 × 10 6 hence significant figure are 2.
90.a.
Total pressure of mixture = P1 + P2
C
=
V1 × P1 V2 × P2 + V V
=
400 × 500 666.6 × 600 + 3000 3000
N
78.b. 79.
14
17
−β −β a) 14 9 X →8 X → 7 N
91.b
Number of neutrons in X are :
= 66.67 + 133.33 = 200 torr The longest wevelength in the balmer series will be :
atomic mass - atomic number = 14 - 9 = 5
1 1 1 = v = R 2 − 2 λ n2 n1
PAGE
12 0.147 × × 100 = 20.245 80.a. % of C = 44 0.2 % of H =
1 1 = 109677× 2 − 2 3 2
2 0.12 × ×100 = 6.666 18 0.2
% of O= 100-(20.045+6.666)=73.29 81.d.
O - H......O
= 109677×
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5 36
= 15232.9 cm-1
N = C - CH 3
CH 3 − C = N | Ni CH 3 − C = N N = C - CH 3
∴ 92.b. 93.c. O || C
CH || CH
C || N
C || O
H -O
1 15232.9
= 6.565 x 10-5 cm = 656.5 nm
O - H...... O
N
λ=
1 ml solution of HCl contains 1.17 gm of HCl
∴
molarity =
1.17 1000 × 36.5 1
= 32.05 O | OH
94.d. 95.a. 96.d.
On increasing the pressure, boiling point increases and the beans are cooked earlier in pressure cooker Only phosphorous can form these compounds PCl 3 ,
P2 O 5 , Ca 3 P2 N can give two NCl 3 and N 2 O 5 B and 82.c.
-H
+
→
Al one each BCl 3 and AlCl 3
After the removal of 97.b.
one proton (H + ) from cyclopentadiene, a pentadienyl
83.a
PAGE 98.c.
anionis obtained. It has a 6π electron system and is, therefore, aromatic. Cl in 2,4 ,6, trinitrochlorobenzene is activated by there
is exothermic. Hence temp willincrease.
99.c.
100.b. CuS and CdS are ppted by H 2S Hydroxide of AI will
NO 2 groupsat o, and p-positions and hence under goes hydrolysis most readily. 84.b. 85.c
By addition of SO 2 equilibrium wil shift to RHS which
pas into the solution inthe form of NaAlO2 being
amphoteric in nature Hence filtrate will give test for sodium and aluminium 101.c. Hair contains amino acids which upon fusion with soda
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Tetrabutyl ammonium chloride is a phase-transfer catalyst.
lime NaOH+CaO) evolve NH 3
86.d.
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102.c
In structre (c), nitrogen has 10 electrons in the valence shall which is not possible.
λ=
103.a. C 5 H10 has 1° degree of unsaturaton since theisomers are acyclic, all of these are alkenes. For writing the isomers, first introduce the double bond at differenct possible positios, and then consider the possibility of branchg in
K .E . =
1 2 mv 2
....(2)
2
K .E . =
monosubstitude (i) and (iv), disubstituted as in (ii), (iii) and (v) and trisubstituted as in (vi). 104.d. 105.d. Magnesium reacts with ethanol to form magnesium ethoxide which then reacts with water toregenerate
1 h m× ; 2 m× λ
i.e., K .E. ∝
ethanol along with Mg (OH) 2 106.a. On the basis of structure of guanine and complementary bases present in them we can say that if the sequence of bases in the strand of DNA is I, then the sequence in the second strand should be II A:T:G:C:T:T:G:A I T : A : C : G : A : A : C : T II
114.
1 h2 × 2 m× λ2
1 (If λ are same) m
a) The contribution of 8 atoms of fcc unit cell is 8 ×
PAGE unit cell is 6 ×
1 =1 8
1 =3. 3
∴number of atoms in fcc unit cell = 1 + 3 = 4
115.
have NO 2 group differently linked to central metal
c) The process of hydration may be enpresssed as :
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O
=
and the contribution of 6 face centered atom shared by 2
107.c. The complex compound [Co(en) 2 NO 2 Cl] Br can
∆H
→ CuSO 4 (aq ) CuSO 4 (S) 1 − 66.5 kJ
or O - N = O
O
∆H
According to Hess’ s law : ∆H = ∆H + ∆H2
MnO 4 − can not be used in aqueous HCl.
∆H = ∆H1 - ∆H2
2SO 3 (g)
= - 66.5 - 11.7 = - 78.2 kJ
2 moles
116.
∆H o = −198 KJ
a) If the ratio of
Reaction is exothermic and moles are decreasing on RHS hence high pressure and low temperature will shift forward reaction. 110.a. Normality of oxalic acid =
=
6.3 ×1000 = 0.4; 63 × 250
or 0.81x
N1V1 = N 2 V2 V1 × 0.1 = 10 × 0.4 ∴ V1 = 40 mL
- 11.7 kJ CuSO4.5H2O
and Cl 2 hence it will oxidise both Fe + + and Cl − hence
109.c. 2SO 2 (g) + O 2 (g)
∆H2
?
108.a. The reduction potential of MnO 4 − is more than Fe
3 moles
...(i)
Now putting the value of v from equation (i) we get,
the alkyl group, C 5 H10 may be
atom -N
h mv
10 11 5 B and 5 B
is x : y then :
10 x + 11y = 10.81 x+y
= 0.19 y or
x 19 = y 81
117.a. CH3 group is electron repelling group ( +1)
PAGE
while - NO2 group is electron attracting group ( - 1) -NO2 attracts electron density comprehensively when it is attached in ortho and para positin on benzene ring. Hence the order of acidic strength is :
111.a. Mn 2+ (5), Fe 2 + (4), Ti 2 + (2) and
Cr 2+ (4), Mn 2+ contains maximum number of
unpaired electrons hence has maximum magnetic moment.
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112.a. Electropositive character and size increases down the group, the ionic character increases. 113.d. de Broglie wavelength
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118.c. 119.b. For M2S salt : KSP = 4S3 = 4 x (3.5 x 10-6)3 = 1.71 x 10-16
6
6
C p −1 3 7 − p 2 p −1 = 6 C1 35 21
= 6 x 35 x 2 = 2916 ≠ 4860 Again when p = 3
We have 4 =| z + iw |2
6
= (z + iw )(z − iw)
⇒
zz + iw z − izw + ww
Also
4 =| z − iw |2 = (z − iw)(z + iw)
⇒
4 = zz − iwz + iwz + w w
C p −1 3 7 − p 2 p −1 = 6 C 2 3 4 2 2
6×5 × 81 × 4 = 4860 2 ×1
=
... (1)
when p = 3, (1) is true. Thus, p = 3
.... (2)
124.c.
(x + y) dx + x dy = 0
. . . (i)
PAGE
Subtracting (2) from (1), we get
dy x+y = − dx x
iw z − iwz + iw z − iwz = 0
⇒
⇒
w(z − z) + w(z − z) = 0
Which is homogeneous .
⇒
(w + w)(z − z) = 0
⇒
w = −w or z = z
Also (i) canbe written as :
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dy 1 + y = −1 dx x
Which is linear differential equation.
If w = - w , then Re w = 0 and also |w| = 1
125.b.
Thus, equation (i) is both homogeneous and linear. We have,
w = ±i
∴
π − 2B A − B+ C 2ac × sin = 2ac sin 2 2
2
Thus
4 =| z − 1 | = (z − 1)(z − 1)
⇒
4 =| z |2 −z − z + 1
⇒
4 = 1 − (z + z) + 1 ; as | z |2 = 1
⇒
(z + z ) = − 2
π − B = 2ac cos B 2
= 2ac sin
⇒ Re z = -1 ⇒ Im z = 0, since |z | = 1
a 2 + c2 − b2 = 2ac 2ac
∴
z = -1
126.d.
Again if z = z , then 1 = |z|2 = z z = zz = z2 Hence, z = 1 or - 1
= a2 + c2 - b2 = c2 + a2 - b2 Sum of the observation = 100 x 40 = 4000
∴
Corrected ΣX = incorrect ΣX - 50 + 60 = 4000 - 50 + 60 = 4010
122.b. Here. p+q=-p
∴
pq = q
if
q= 0
if
then q ≠0 p = 1 or 0
Hence, 123.b.
and
then
Corrected mean =
p = 0.
PAGE
p = 1 and q = - 2
=
= 6 C p −1 (3) 6 − p +1 (2x )
p −1
= 6 C p −1 3 7 − p 2 p −1 x p −1 As per question
127.a.
Corrected Σ X 100
4010 = 40.10 100
pth term in the expansion of ( 3 + 2x)6
∴
..... (1)
By trial, when p = 2, we have
K × W × 1000 2.16 × 0.11 × 1000 120.b. M = b = = 158.4 ∆Tb × w 0.1 × 1.5
121.c.
C p −1 3 7 − p 2 p −1 = 4860
→
→
→
Given that | a | = | b | = | c | = 1 and →
→
→
[ a , b, c ] = 0
www.aieeepage.com ∴ [2 a − b , 2 b − c , 2 c − a ] →
→
→
→
→
→
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→
→
→
→
→
∴
→
= (2 a − b ) . [(2 b − c ) × (2 c − a )] →
→
→ →
→ →
→
133.d.
→
→ →
→ →
→ → →
⇒ x ≠ -1, -2
→ →
Hence, domain = (-3,
(2 a − b ) . ( 4 b × c − 2 b × a + c × a ) → → →
→ → →
For domain, x + 3 > 0 ⇒ x > - 3 and x2 + 3x + 2 ≠ 0 ⇒ (x + 1)(x + 2) ≠ 0
→ →
= (2 a − b ) . [ 4 b × c − 2 b × a − 2 c × c + c × a ] →
the smallest value of n is 7.
134.b.
∞ )\{-1, -2}
Required Area is
→ → →
= 2 × 4[ a , b , c ] − 4 [ a , b , a ] + 2[ a , c , a ] − [ b , c , a ] → → →
→ → →
→ → →
8 [ a , b, c ] − [ a , b, c ] = 7 [ a , b, c ] = 0 128.c. φ(x) = f -1 (x); ⇒ x = fφ( x)) ⇒ 1 = φ' (φ( x ))
⇒ φ'( x) = f ' ( x) =
ABCD = (AB)2
1 f ' (φ(x )) 1
1+ x
5
= ( 2) 2
=2
PAGE 135.d.
⇒ f ' (φ( x)) =
1
1 + (φ( x ))
Equation of a plane through (-2, -2, 2) is given by
5
a(x + 2) + b(y + 2) + c (z - 2) = 0
It contains the line joining the points (1, 1, 1) and (1, -1, 2), so these points also lie in the plane
1 = 1 + (φ( x ))5 or φ' ( x) = f' (φ( x ))
∴ 3a + 3b - c = 0; 3a + b+ 0. c = 0
a b c = =r ∴ = www.aieeepage.com 1 −3 −6
129.c. x 2 + y 2 + z 2 − 2 x − 2 y − 2 z + λ = 0 is a family of concentric sphere. If sphere through (1, 1, 1),
or equation of the plane is
then λ = 3 130.d. f = { (x , u), (y, v), (z, v)} is the given mapping from A = { x, y, z} to B = {u, v, w}
x - 3y - 6z + 8 = 0 136.a.
Thus f is many one into
y2 + z2 = 0 ⇒ y = 0, z = 0 In three dimensional space, y = 0 and z = 0, represents xz - plane xy - plane respectively, whose intersection is x-axis.
Clearly f(y) = f(z) = v, so f is many one and range of f = {u, v} ⊆ B, so it is into. 131.d.
(say)
137.b.
Here, augmented matrix
Here if velocity of current is u, then velocity of man = 2u
1 2 −3 1 0 k+2 3 =0 0 2k + 1 1 2 1 2 = 0 2k + 1 0
0
−3 1
1 2
k+2
3
PAGE ∴
interchanging R2 and R3
Resultant velocity is along OC, ⊥ to OA Resolving along OA,we get
This shows that the system of equations will be inconsistent if k + 2 = 0 i.e., k = - 2 132.b.
Distinct n digit numbers which can be formed using digits 2, 5 and 7 = 3 x 3 x 3 ....... upto n times = 3 n We have to find n so that 3 ≥ 900 n
u + 2u cos θ = 0 ⇒ cos θ = -1/2 ⇒
θ = 120o
138.c. Since log e (log e x) is defined if x > 1 139.c. For both the roots to be positive the condition are
www.aieeepage.com 2a − 1 >0 D ≥ 0, f (0) > 0,
⇒
3 n ≥ 32 × 100 ⇒ 3 n −2 ≥ 100
⇒
n−2≥5
⇒
OA = u, OB = 2u
2
n≥7
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143.d. Equation of a plane through
2+ 3 ⇒ a ∈ , ∞ 2
(2, - 3, 1) is a (x -2 ) + b (y + 3)+ c (z - 1) = 0
140.d. Ap = h cot α, BP = h cot β
-3a + 4b - 8c = 0 so
Q
h α
β
B
h cot β
h cot α
B
Putting these values of a, b, c in (1), equation of the required plane is 4x + 7y + 2z + 11 = 0
PAGE
100
15 =0 2
and
C
h cot β P (Horizontal plane)
2
144.d. 3x + 4 y −
100m
c −c = www.aieeepage.com a +b 2
2
cot α + cot β
3x+4y - 9= 0
since these lines are parallel, so distance between these lines
1002 = h 2 cot 2 α + h 2 cot 2 β h=
.... (3)
a b c = = =λ 4 7 2
P
(Vertical plane) A 100m
a - 2b + 5c = 0
Solving (2) and (3), we get
h
p
h cot α
.... (2)
It is perp. to the plane x - 2y + 5z + 1 = 0
Q
A
.... (1)
It passes through ( - 1, 1, - 7) , so
1
2
2
15 −9 −− 2 = 2 3 + 42
141.b. a1 + a2 n = a 2 + a2 n − 1 = an + an + 1 = k (say)
a1 − a 2 a n − a n +1 + ... + Expression = k a − a a n − a n +1 1 2 =
=
k ( a1 − a n +1 ), where d = common difference −d
k −d
a1 − an +1
= (a1 + a 2n ).
a1 + an +1
=
145.c.
Limit = xlim →∞
− nd = lim
− d ( a1 + a n +1 )
x →∞
142.b. We have 13
13
n =1
n =1
x+
x−x
x+
x +
x
1 x = 1 2 x 1 + + 1 x
146.a. Here only the variate values are changed.
n n +1 n ∑ (i + i ) = ∑ (1 + i)i
Hence the only probability of xi and f ( xi ) is the same pi.
= (1 + i)[ i + i 2 + i 3 + ..... + i13 ]
i(1 − i13 ) = (1 + i) 1 − i i(1 − i) = (1 + i) 1− i
3 3 2 = 2 = 3 5 10 9 + 16 −
PAGE
147.a. Given n p = 9 and
npq = 6
⇒ n p = 9 and n p q = 6
⇒q=
6 2 2 1 = ⇒ p = 1− q = 1− = 9 3 3 3
and np = 9
www.aieeepage.com
as i13 = (i4 )3 .i = 1. i = i
⇒n=
= ( 1 + i) i = i - 1
9 = 27. Hence n = 27 and p = 1 p 3
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154.d. Let two numbers be x, 3 - x
148.a. x18 = y 21 = z 28
∴ P = x(3 − x) 2 ⇒
⇒ 18 log x = 211 log y = 28 log z
Now3, 3 log y x, 3 log z y, 7 log x z ⇒ 3, 3
∴
log x log y log z 21 28 18 ,3 ,7 , ⇒ 3, 3 ,3 ,7 log y log z log x 18 21 28
⇒
dx 2
2
1 − x dy + 1 − y dx = 0 dy
dx
+
1− y2
=0
1− x2
dx2
d 2P
(-4, 5) to (3, -2) and that is when a = 7 2 150.b. Given
d2p
= −6 < 0 x =1
∴ P is max. at x = 1 and maximum value is
PAGE 1(3 − 1) 2 = 4
integrating, we get
sin −1 y + sin −1 x = c
155.b.
(1+y2) dx - xy dy = 0
⇒
151.a. Required left hand derivative
= f ( k − 0) = lim h →0
dx y dy = x 1 + y2
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f ( k − h ) − f ( k) −h
= lim h →0
( k − 1) sin π( k − h ) − k sin kπ −h
= lim h →0
( k − 1) (−1) k −1 sin πh − 0 −h
= ( k − 1)( −1) k π lim h →0
= (3 − x)(−3) + (3 − 3x)(−1) = 6 x − 12
dp = 0 ⇒ x = 3, I dx But x = 3 is not possible
149.b. As the line through (-4, 5) varies, the distance a of the line from (3, -2) also varies. The minimum value of a is when the line is perpendicular to the line joining
2
dp = (3 − x)(3 − 3x) dx
and
⇒ 3, 3.5,4.4.5 (anA.P).
dp = −2 x(3 − x) + (3 − x )2 dx
sin πh πh
1 log (1 + y 2 ) + log c 2
⇒
log | x | =
⇒
| x | = c 1+ y2
When x = 1, y = 0 so we get c = 1 ∴ solutin is x = 1 + y 2 or
x2 - y 2 = 1
156.a. Here X = 2, Y = 1 tan θ = Y/X = 1/2
= ( k − 1)(−1) k π × 1 = (−1) k ( k − 1)π 152.b. Here,
10 − ( r −1)
x Tr = 10 C r −1 3 11− r
1 = 10 C r −1 3
2 − 2 x
r −1
PAGE
Since, rth term contains x4, so that 13 - 3r = 4 or 153.
Equation of the line of action of the resultant is y = ( tan θ) (x) i.e.
(− 2 )r −1 x13−3r
y = (1/2) x i.e. y - 2x = 0
157.b. We have
ˆi ˆj kˆ a × b = 2 1 − 2 = 2ˆi − 2 ˆj + kˆ 1 1 0
3r = 9 ⇒ r = 3 ⇒
3 25 Probability of drawing a red ball = 5
b)Probability of drawing a white ball =
The initial ball drawn would be either white or red. ∴ probability of drawing a white ball in second attempt without replacement
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∴a × b = 1 = 4 + 4 + 1 = 3 2
Again, since c − a = 8
3 2 2 2 1 = × + × = 5 4 5 4 2
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12
2
∵ 0 < α, β < 90° ⇒ 0 < 2α < 180°
2
c + a − 2c . a = 8
⇒
2
c + 9 − 2 c = 8;
⇒
161.b. Since α, β, γ are the roots of the equation
as a = 4 + 1 + 4 = 9 and c.a = a.c = c 2
x 3 + mx 2 + 3 x + m = 0; ∴ α + β + γ = -m
αβ + βγ + γα = 3; αβγ = -m
2
⇒
c − 2 c +1 = 0
Now, If tan −1α = θ1; tan −1 β = θ 2 ; tan −1 γ = θ3
⇒
{ c −1 } = 0 ⇒ c = 1
tan(θ1 + θ 2 + θ)3 ) =
2
1 3 = (3)(1) = 2 2 158. c. .Let
2
2
a ∆ = (a + 1) 2 (a − 1) 2
=
⇒ tan−1 α + tan-1 β + tan-1γ = nπ
c (c + 1) 2 (c − 1) 2
162.d. P = (cos θ, sinθ ), Q = {(cos (α - θ), sin (α - θ),}
www.aieeepage.com y = tan
Apply R 2 → R 2 − R 3 and take 4 common a2 ∆=4 a (a − 1) 2
b2 b (b − 1) 2
c2 c (c − 1) 2
(α + β + γ) −m − (−m) = =0 1 − (αβ + βγ + γα) 1− 3
∴ θ1 + θ2 + θ3 = nπ
2
b (b + 1) 2 (b − 1) 2
b2
c2
∆=4 a 1
b 1
c = 4(a − b)(b − c)(c − a) 1
α x 2
sin(α - θ) + sin θ α cos( α - θ) + cosθ = tan 2 2 2
Apply R 3 → R 3 + 2R 2 − R 1
a2
P θ
θ/2
-θ
Now given ∆ = 0 ⇒ 4(a-b)(b-c)(c-a)=0 Clearly, atleast one of the factors must be zero. Hence the triangle must be isosceles. ALTERNATE:
α α - 2θ 2 sin cos α α 2 α - 2θ 2 = = tan .2 cos cos 2 2 1 2
If a = b then ∆ =0 as C1 and C2 become identical. Similarly if b = c or c = a, ∆ = 0. 159.a. The function f ( x) =
x 3
x + 16
is increasing in [0, 1] = 2 sin
∴ Min. value of f(x) is f(0) = 0 and Max. value of f (x ) is f (1) =
1 17
x 3
x + 16
dx ≤
α α - 2θ α α α - 2θ cos = tan 2 cos cos 2 2 2 2 2
PAGE
It means, it is the reflection of y = tan
Since m(b − a) ≤ ∫ba f ( x)dx ≤ M (b − a) 0 ≤ ∫10
Σ tan θ1 − tan θ1 tan θ 2 tanθ3 1 − Σ tan θ1 tan θ 2
PAGE
Now, (a × b) × c = a × b c sin 30°
1 17
160.b. tan 2α = tan {(α + β) + (α - β)} =
1
∴ 2α = 135° and sin 2α =
2
tan(α + β) + tan (α − β) = −1 1 − tan(α + β)tan (α − β)
α x 2
163.b. Direction rations of the given diagonal are < a - 0 a - 0, a - 0 > or < 1, 1, 1 > where ‘a’ is the the edge of cube. n 164.d. s = kt ⇒ v =
www.aieeepage.com f =
ds = nkt n −1 dt
dv = n( n − 1) kt n − 2 ; ⇒ ( n − 1)v 2 = ( n − 1) n 2 k 2 t 2 n − 2 dt
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13
⇒ nfs = n. nk (n − 1)t n − 2kt n ;
a ≤ 3 and x = a ± (3 − a )
= n 2 ( n − 1) k 2t 2 n − 2
Since both the roots of the given equation are less than 3, so the largest of the two roots is less than 3.
⇒ ( n − 1)v 2 − nfs
=±
a×b | a×b |
4 i − j − 3k
=±
4 2 + (−1) 2 + (−3) 2
1
Now, k. ±
26
=
(σ ) =
26
1
(4 i − j − 3k) = ±
250 = 25 = 5 10
26
(4 i − j − 3k)
(−3)
26
−1
(4 i − j − 3k)
a
2
4R
2
+
b
2
4R
2
+
c
2
n
n
7π / 6
π
π
11π / 6
7π / 6
=
∫
7π / 6
π
[2 sin x]dx + (−1)dx +
∑ r =0
n
Cr n
= na n − ∑
r =0
n
∑
−.
r =0
Cr
1 4R 2
(16 R 2 ) = 4 =− 172.a.
11π / 6
11π / 6
∫
7π / 6
[2 sin x]dx
(−2)dx +
∫
2π
11π / 6
(−1)dx
π 8π π 5π − − =− 6 6 6 3
From the first relation al + bm n = − c Putting it in second relation
PAGE
(n − r) ; as n C r = n C n − r n C n −r
= na n − S ∴ 2S = na n or S =
al + bm al + bm fm − + hlm = 0 + gl − c c or afml + bfm2 + agl2 + bglm - chlm = 0
1 na n 2
or ag
a)Solution : The given equation can be written as (x a)2 = 3 - a
a ≤ 3 and x = a ± (3 − a )
2π
11π + (−1) 2π − 6
n−r n
∫
11π 7π 7π − = (−1) − π + (−2) 6 6 6
r n−r = ∑ n C r r =0 C r
1
⇒a<2
2n
2π
∫
4R 2
n
r =0
169.
(3 − a) > 1
a 2 + b 2 + c 2 = 16R 2
S= ∑
(3 − a ) > 1
∴ (1 + i )2 n + (1 − i )2 n = z + z = 2 Re( z ), a pure real number
a2 + b2 + c2 = 2(2 R ) 2 2
n
=n
a < 3 and
∴ z = (1 + i) 2 n = (1 + i ) 2n = (1 − i) 2n
of
(a 2 + b 2 + c 2 )
∴ sin 2 A + sin 2 B + sin 2 C = 168.c.
⇒
171.a.
∴ sin 2 A + sin 2 B + sin 2 C =
⇒
a < 3 and 1 − (3 − a) < 0
www.aieeepage.com ∫ [2 sin x] dx =∫ [2 sin x]dx +
∴
Given,
⇒
PAGE
where R is the circum radius
4R
(3 − a ) (1 − (3 − a ) ) < 0
∴a < 2 and a < 3
a b c = = = 2 R, ∆ABC, sin A sin B sin C
2
⇒
⇒ 3 − a > 1 or a < 2
circumcircle
1
(3 − a) − (3 − a) < 0
170.b. Let z = (1 + i )
a
=
⇒
∵
and hence coefficient of variation
5 σ × 100= × 100 = 10 mean 50
167.c. In
1
=±
for acute angle the unit vector =
166.c S.D.
a + (3 − a) < 3
∴
165.a. The unit vector
l2
m2
+
www.aieeepage.com
l (af + bg - ch) + bf = 0 . . .(i) m
Now, if l1, m1, n1 and l2, m2, n2 be direction cosines of two lines, then from (i)
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14
l1 l2 bf = m1 m 2 ag
3m +
1 m
=3
1+ m2
l1 l2 , sin ce roots of (1) are m1 m 2
⇒
9m4 + 1 + 6m2 = 9m2 ( 1 + m2)
l1 l2 m1m2 = or f /a g /b
⇒
3m2 = 1
y= 175.c.
l1 l2 m1m 2 n1n 2 ∴ f / a = g / b = h / c = q (say)
1 3
vertices of the tetrahedron are (0, 0, 0), (6, 0, 0), (0, -4, 0), (0, 0, 3).
PAGE
0 0 0 1 16 0 0 1 ∴volume = 6 0 −4 0 1 0
i.e. if (f/a) q + (g/b) q + (h/c) q = 0 i.e. if f/a + g/b + h/c = 0 173.
d)Let AB and AC represent the forces P and Q, then the diagonal AD will represent their resultant R. (say)
x2
176.a. Let eq. of hyperbola be
a2
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∴
R divide ∠BAC in the ratio 1 : 2,
∴
let ∠BAD = θ and ∠CAD = 2θ
3
x + 3 or 3y = x + 3
We know that the lines are perpendicular if l1 l2 + m1 m2 + n1 n2 = 0
1
But m > 0, then equation of common tangent is
Similarly elimination of l will yield
m 1m 2 n 1 n 2 = g/b h/c
m =±
⇒
0
−
y2 b2
3 1
= I centre is (0, 0),
vertex (a, 0). focus (ae, 0).
e = eccentricity =
In ∆ABD , by sine rule
a=
P Q R = = sin 2θ sin θ sin(π − 3θ)
1 2 a + b 2 . It is given that a
2ae 3
⇒ a2 =
4 2 2 4 2 a e = (a + b 2 ) 9 9 ⇒ 9a 2 = 4a 2 + 4b 2 ;
⇒ b2 =
∴
P/2A = cos θ
and
R = Q sin 3 θ /sin θ = Q(3 - 2 sin2 θ )
∴ Eq. of hyperbola is
= Q[3 - 4(1 - cos2 θ )] = Q(-1 + 4 cos2 θ ),
a2
−
y2 5a 2 4
=1
PAGE
174.c. Let the common tangent to the parabola y2 = 4x be
1 m
x2
⇒ 5x 2 − 4y 2 = 5a 2
= Q[-1 + 4P2 / 4Q2] = (P2 - Q2) Q
y = mx +
5 2 a 4
It should also touch the circle (x- 3)2 + y2 = 9
177.d. Let W descends and ascends. W' Let f be the acceleration of the system. Since the pulley ascends with an acceleration f, the actual acceleration of W and W ' = f - g. (downwards) and f + g (upwards) respectively. Let T be the tension in the string. Then W g - T = W (f - g) and
T − W ' g = W ( f + g) www.aieeepage.com
whose centre is at (3, 0) and radius = 3, then
⇒
g-
T = f −g W
and
T −g = f +g W'
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subtracting,
=T =
we
get
T T + − 2 g = 2g W W'
4 W W' kg wt W + W'
178.b. C1C2 = C1A − C2A
< R − r (∵ C2 A > r ∴ −C2 A < −r ) A C2
C1
179.b. No. of large cages = m - n No. of lalrge animals = p
PAGE
No. of ways of arranging large animals = m − nPp
No. of remaining cages = m - p ∴ No. of ways of arranging remaining m - p animlas = m − pPm − p
m−n Pp )(m − pPm − p ) ∴ Total no. of ways = (
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180.b. x2 = x1r , x3 = x1r 2 , y 2 = y1r ,y3 = y1r 2
y1r 2 − y1r 2
x1r − x1r
=
y1r − y1 ⇒ points are collinear x1r − x1
PAGE
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16
PHYSICS ANALYSIS Sl.No.
Unit Name
Question No.
Correct
Wrong
1. 2. 3. 4. 5. 6. 7.
PAGE
8. 9. 10. 11. 12.
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13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.
PAGE
25. 26. 27. 28.
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29. 30.
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17